Problem: The lifespans of lizards in a particular zoo are normally distributed. The average lizard lives $2.7$ years; the standard deviation is $0.5$ years. Use the empirical rule (68-95-99.7%) to estimate the probability of a lizard living between $2.2$ and $4.2$ years.
$2.7$ $2.2$ $3.2$ $1.7$ $3.7$ $1.2$ $4.2$ $99.7\%$ $68\%$ $15.85\%$ $15.85\%$ We know the lifespans are normally distributed with an average lifespan of $2.7$ years. We know the standard deviation is $0.5$ years, so one standard deviation below the mean is $2.2$ years and one standard deviation above the mean is $3.2$ years. Two standard deviations below the mean is $1.7$ years and two standard deviations above the mean is $3.7$ years. Three standard deviations below the mean is $1.2$ years and three standard deviations above the mean is $4.2$ years. We are interested in the probability of a lizard living between $2.2$ and $4.2$ years. The empirical rule (or the 68-95-99.7 rule) tells us that $99.7\%$ of the lizards will have lifespans within 3 standard deviations of the average lifespan. It also tells us that $68\%$ of the lizards will have lifespans within 1 standard deviation of the mean. The probability of a particular lizard living between $2.2$ and $4.2$ years is ${68\%} + \color{orange}{15.85\%}$, or $83.85\%$.